Câu 1:
$f(x)=\begin{cases}\dfrac{2-\sqrt{x+3}}{x^2-1}\,\,\,\text{ khi }x\ne 1\\\dfrac{1}{4}+2m\,\,\,\text{ khi }x=1\end{cases}\\D=[-3;+\infty)/\{\pm1\}\\\text{Hàm số liên tục trên }[-3;-1),\,(-1;1)\text{ và }(1;+\infty)\\\text{Xét tại }x=1: \\\lim\limits_{x\to 1^+}\dfrac{2-\sqrt{x+3}}{x^2-1}=\lim\limits_{x\to 1^+}\dfrac{4-x-3}{(x-1)(x+1)(2+\sqrt{x+3})}=\lim\limits_{x\to 1^+}\dfrac{-(x-1)}{(x-1)(x+1)(2+\sqrt{x+3})}\\=\lim\limits_{x\to 1^+}\dfrac{-1}{(x+1)(2+\sqrt{x+3})}=\dfrac{-1}{(1+1)(2+\sqrt{1+3})}=-\dfrac{1}{8}\\\lim\limits_{x\to 1^-}\dfrac{2-\sqrt{x+3}}{x^2-1}=\lim\limits_{x\to 1^-}\dfrac{4-x-3}{(x-1)(x+1)(2+\sqrt{x+3})}=\lim\limits_{x\to 1^-}\dfrac{-(x-1)}{(x-1)(x+1)(2+\sqrt{x+3})}\\=\lim\limits_{x\to 1^-}\dfrac{-1}{(x+1)(2+\sqrt{x+3})}=\dfrac{-1}{(1+1)(2+\sqrt{1+3})}=-\dfrac{1}{8}\\\text{Để hàm số liên tục tại }x=1\\\to \lim\limits_{x\to 1^+}=\lim\limits_{x\to 1^-}=f(1)\\\to -\dfrac{1}{8}=\dfrac{1}{4}+2m\\\to m=-\dfrac{3}{16}$
Câu 2:
$y=\dfrac{x^3}{3}+3x^2-2\\\to y'=x^2+6x\\\to x^2+6x=-9\text{ (vì }y'=k)\\\to (x+3)^2=0\\\to x=-3\\\to y(-3)=\dfrac{(-3)^3}{3}+3.(-3)^2-2=16\\\to \text{Phương trình tiếp tuyến: }y=-9(x+3)+16=-9x-11$
